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Chapter 7: Problem 42

Graph each ellipse and give the location of its foci. $$\frac{(x-3)^{2}}{9}+\frac{(y+1)^{2}}{16}=1$$

Short Answer

Expert verified
The ellipse has its center at (3, -1), semi-major axis of length 4, and semi-minor axis of length 3. The foci of the ellipse are at (3, -1卤sqrt(7)).

Step by step solution

01

Identify the Center, and Semi-major and Semi-minor Axes

The center of the ellipse is the point (h, k), which here is (3, -1). Since the denominator of the x-term is less than that of the y-term, the semi-major axis is along the y-axis and has length \(a=\sqrt{16}=4\) and the semi-minor axis is along the x-axis and has length \(b=\sqrt{9}=3\).
02

Draw the Ellipse

Start by plotting the center of the ellipse at (3, -1). Sketch the ellipse by plotting points a units above and below the center, and b units to the right and left of the center. This gives points at (3, 3), (3, -5), (0, -1), and (6, -1). Draw a smooth curve through these four points to finish the ellipse.
03

Find the Foci

The foci of an ellipse are located on the major axis, c units from the center, where \(c^2=a^2-b^2\). Here, \(c=\sqrt{16-9}=\sqrt{7}\). Therefore, the foci are at (3, -1卤sqrt(7)).

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Most popular questions from this chapter

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{64}+\frac{y^{2}}{100}=1$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$$

Convert each equation to standard form by completing the square on \(x\) or \(y .\) Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola. $$ x^{2}-2 x-4 y+9=0 $$

What is a hyperbola?

Graph each ellipse and give the location of its foci. $$\frac{(x-1)^{2}}{2}+\frac{(y+3)^{2}}{5}=1$$
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